# Will car burn same amount of fuel with equal acceleration $(ΔV)$ but at different speeds if resistance forces don't exist?

Let's imagine that resistance forces(aerodrag, rolling resistance, internal friction, etc.) doesn't exist.

1. Will a car consume the same amount of fuel when accelerating from $$0-100km/h$$ and when accelerating from $$100-200km/h$$?

2. Formula for kinetic energy is $$E=(1/2) mv^2$$, so with speed rise kinetic energy. Does that mean that the car uses more fuel (higher power) when travelling at constant $$200km/h$$ than when travelling at constant $$100km/h$$?

(hmm.. if no resistance forces, car cannot travel at constant speed, he will accelerate towards speed of light..hmm)

1. Since KE depends on velocity squared, each additional km/h requires more energy than the last. Suppose you have a car that rolls down a hill, converting potential energy into kinetic energy, and achieves some speed X at the bottom. If you double the height of the hill and double the kinetic energy gained, the speed of the car at the bottom is less than 2X. Accelerating from 0-X required some amount of energy, but accelerating from X-2X requires more than that.

2. In absence of resistive forces, it takes no energy whatsoever to maintain a constant speed. A spaceship floating through empty space will float along at a constant speed forever, unless something acts to change its velocity. If there is no drag or rolling resistance, a car can coast at a constant speed with the engine off and consume no fuel at all. In reality, fuel is required to negate the resistive forces, which tend to increase with speed, so a car is typically less fuel efficient at 200km/h than at 100km/h, since there is more drag.

• But if car has constant thrust rocket engine ,then acceleration is same from 0-100km/h and 100-200km/h ,and it will burn same amount of fuel in both case? if I neglect fuel mass loss over time Nov 4 '21 at 16:39
• @JurgenM $W = Fd$, so as the car speeds up, it covers a greater $d$ per unit time, meaning that a constant-force/acceleration rocket does increasingly more work on the car the faster it goes. You'd think that the rocket doing more work would mean more fuel consumed (the energy must come from somewhere), but that's not the case - the total energy of the rocket plus the exhaust works out so that the total KE increases at whatever constant rate fuel is burned. This is called the Oberth Effect, and it seems like a paradoxical way to get energy "for free" until you consider the exhaust. Nov 4 '21 at 16:57
• @JurgenM The Oberth Effect has interesting implications for space travel - a fixed amount of fuel can do different amounts of work on a rocket depending on how fast it's going! It's most efficient to do a burn when at orbital periapsis when your velocity is highest, as that will result in the highest gain of KE for the rocket. Nov 4 '21 at 17:01
• So rocket has same acceleration from 0-100 and 100-200,or even faster acceleration from 100-200 if we include mass loss over time?And consume same amount of fuel in both case? So everything opposite from car? Nov 4 '21 at 17:12
• @JurgenM $F=ma$, and a constant-thrust rocket by definition has constant $F$ - with constant $m$ you get constant $a$, and with decreasing $m$ you get increasing $a$, all for constant fuel consumption. As the fuel itself accelerates, its own total energy rises, allowing it to do more work. I think a key difference is that in a rocket, speed of the rocket and exhaust are both measured relative to some other frame, while in the car, speed is measured relative to the "exhaust", which is the road being pushed backwards. Nov 4 '21 at 18:09

As per Newton's second law and in the absence of any loss functions (as you specified) the car power requirement will be independent of driving speed and zero, despite the fact that at higher speed the car has more kinetic energy, acc. $$K=\frac12 mv^2$$.

Power is only needed when the car accelerates because then energy has to be added to increase the kinetic energy K.

The energy needed to accelerate the car from, say $$v_1$$ to $$v_2$$, is $$\Delta K$$, or:

$$\Delta K=K_2-K_1=\frac12 mv_2^2-\frac12 mv_1^2$$

$$\Delta K=\frac12 m (v_2^2-v_1^2)$$

You can now insert your own speeds to get that part of your answer.

(hmm if no resistance forces,car can not travel at constant speed,he will accelerate towards speed of light..hmm)

Yes, in the absence of any losses the powered car will continue to accelerate until relativistic effects set in. In the Real World that doesn't happen of course because losses are always substantial.

Will a car consume the same amount of fuel when accelerating from 0−100km/h and when accelerating from 100−200km/h?

No. It is easier to understand supposing a constant acceleration. In this case: $$V_2^2 - V_1^2 = 2ad$$ where $$d$$ is the traveled distance.
$$d$$ is greater when going from $$100km/h$$ to $$200 km/h$$. As Work is $$w = Fd$$, and $$F$$ is constant, $$w$$ is greater for the second range of velocities.

Formula for kinetic energy is E=(1/2)mv2, so with speed rise kinetic energy. Does that mean that the car uses more fuel (higher power) when travelling at constant 200km/h than when travelling at constant 100km/h?

If travelling without dissipative forces, no force is required.

Burn more fuel is not the same as more power, power is the rate at which energy is being used up/ rate at which work is done on an object

( explanation about why fuel is needed to maintain speed at all)The rate at which work is being done and therefore more fuel burning) on an object given resistance would be higher for an object traveling at higher speeds, due to the fact that given a constant force (that balances out the resistance force thus resulting in a constant velocity) the rate at which work being done on an object is $$F \cdot V$$. meaning for something that has a higher velocity,more energy is transfered from chemical energy to mechanical energy to overcome the resistive forces ( Note the F used here is the same force that the resistance has at a certain speed) ( and this formula represents the work done by the engine on the car, not the combined work done on the car, which is zero as its a constant speed)

This is obviously not the case if at a constant speed without resistance terms as the F needed to maintain the constant speed is zero.